uva 10706 Number Sequence（找规律）

                                      uva 10706 Number Sequence

A single positive integer iis given. Write a program to find the digit located in the positioniin the sequence of number groups S₁S₂…S_k. Each groupS_kconsists of a sequence of positive integer numbers ranging from1 to k, written one after another. For example, the first80 digits of the sequence are as follows:

11212312341234512345612345671234567812345678912345678910123456789101112345678910

Input

The first line of the input file contains a single integer t (1 <=t <=25), the number of test cases, followed by one line for each test case. The line for a test case contains the single integeri (1 <=i <=2147483647)

 

Output

There should be one output line per test case containing the digit located in the positioni.

 

Sample Input                           Output for Sample Input

2                                                                                                                  

8

3

2                                                                           

2                                                                                                       

题目大意：有一个数字序列  1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011……

给出位数i，求该数字序列第i位为什么数字。

解题思路：具体看注释。

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;int main() {int T;scanf("%d", &T);while (T--) {int n;scanf("%d", &n);long long temp, move = 0, cnt = 0, dig = 0;while (move < n) {cnt++; //标记n在哪一组数据中temp = cnt;while (temp) {dig++; //dig为每一组数据的长度 1)1, 2)12, 3)123,...10)12345678910temp /= 10;}move += dig; //move为 1， 3， 6， 10， 15， 21， 28， 36... cnt组数据的长度}temp = cnt; //cnt组的最后一个数为cntwhile (move != n) { //n为所求位数，move为cnt组数据总位数move--;temp /= 10; //当数据大于一位时， 通过此处使得数据和位数获得一致if (!temp) {cnt--;temp = cnt;}}printf("%lld\n", temp % 10);}}